Enumerative combinatorics deals with finite sets and their cardinalities. In other words, a typical problem of enumerative combinatorics is to find the number of ways a certain pattern can be formed.
In the first part of our course we will be dealing with elementary combinatorial objects and notions: permutations, combinations, compositions, Fibonacci and Catalan numbers etc. In the second part of the course we introduce the notion of generating functions and use it to study recurrence relations and partition numbers.
The course is mostly self-contained. However, some acquaintance with basic linear algebra and analysis (including Taylor series expansion) may be very helpful.
Do you have technical problems? Write to us: coursera@hse.ru

Our final lecture is devoted to the so-called "q-analogues" of various combinatorial notions and identities. As a general principle, we replace identities with numbers by identities with polynomials in a certain variable, usually denoted by q, that return the original statement as q tends to 1. This approach turns out to be extremely useful in various branches of mathematics, from number theory to representation theory.

Impartido por:

Evgeny Smirnov

Associate Professor

Transcripción

[MUSIC] Okay, now, we will define the main character of this lecture, q-binomial coefficients. Take two non-negative integers, m and n. And take a rectangle of size m times n. We can consider the generating function for young diagrams fitting inside this rectangle, just as in this example, where the weight of a young diagram is the number of its boxes. So we take, q to the power, weight of lambda for all lambdas inside this rectangle. What we get is called the q by number of coefficient, m + n choose n. And it is denoted like this, [m + n n] q in square brackets. So this also called Gaussian binomial coefficient. Because it appeared in some number theory problems considered by Gauss. Okay, so this sum is finite, so this thing is a polynomial in q. And what is its degree? So the degree is equal to the maximal weight of a young diagram, namely, to the whole rectangle, m times n. So the degree of this polynomial, Is mn. And this polynomial is monic because the young diagram of such a weight is unique. Its lowest term is equal to one because there is only one diagram with weight zero. So there is one minimal and one maximal young diagram. And you can also see that there is only one sub maximal and one sub minimal young diagram. So the terms in front of q and mn- 1 are also equal to 1. So this thing is 1 + q + etc + q to the power mn- 1 + q to the power mn. Okay, so we already have this example that 4 choose 2 is equal to 1 + q + twice q squared + q cubed + q to the fourth. Okay, let us consider one more, Easy and important case. What happens if m = 1, and n arbitrary? In this case, Our rectangle is a rule of n boxes. And there are n + 1 young diagram fitting inside this rectangle. There are rows of arbitrary length from zero to n. So m + 1 choose 1 sub q, or sometimes, we will omit this q and just write these in square brackets. This = 1 + q + q squared + etc + q to the power n. And this is a geometric series with common ratio q. So this is 1- q to the power n + 1 divided by 1- q. Okay, so let us list some properties of q-binomial coefficients. First of all, what happens if we take this polynomial and evaluate it in q equal to one? M + n choose n sub q evaluated in q = 1 equals just the ordinary binomial coefficient m + n choose n. Well, indeed, when q = 1, each young diagram is counted with the weight 1, so we just compute their number. So the value of this polynomial equals to the number of young diagrams. This is m + n choose n, as we discussed in the beginning of our lecture. And then, what else? Q-binomial coefficients are symmetric, so m + n choose n, = m + n choose n, just as ordinary binomial coefficients. Why is it so? Because every rectangle size, m times n, can be reflected with respect to its main diagonal. And thus, it will give us a rectangle with m columns and n rows. And there is a bijection between young diagrams in this rectangle and in this one, and this bijection preserves weights. Okay, this means that in this aspect, q n m coefficient behave very much like the ordinary ones. They are all symmetric. And what we'll do now is we are going to construct an inductive rule of obtaining, The q-binomial coefficients for given n + n from the previous ones. Just as it happens with the Pascal triangle. [MUSIC]